Problem: The equation of a circle $C$ is $x^2+y^2-4x-4y-28 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-4x) + (y^2-4y) = 28$ $(x^2-4x+4) + (y^2-4y+4) = 28 + 4 + 4$ $(x-2)^{2} + (y-2)^{2} = 36 = 6^2$ Thus, $(h, k) = (2, 2)$ and $r = 6$.